# 19. 删除链表的倒数第 N 个结点 - LeetCode Python/Java/C++/JS/C#/Go/Ruby 题解
访问原文链接：[19. 删除链表的倒数第 N 个结点 - LeetCode Python/Java/C++/JS/C#/Go/Ruby 题解](https://leetcode.to/zh/leetcode/19-remove-nth-node-from-end-of-list)，体验更佳！
力扣链接：[19. 删除链表的倒数第 N 个结点](https://leetcode.cn/problems/remove-nth-node-from-end-of-list), 难度等级：**中等**。
## LeetCode “19. 删除链表的倒数第 N 个结点”问题描述
给你一个链表，删除链表的倒数第 `n` 个结点，并且返回链表的头结点。
### [示例 1]
**输入**: `head = [1,2,3,4,5], n = 2`
**输出**: `[1,2,3,5]`
### [示例 2]
**输入**: `head = [1], n = 1`
**输出**: `[]`
### [示例 3]
**输入**: `head = [1,2], n = 1`
**输出**: `[1]`
### [约束]
- 链表中结点的数目为 `sz`
- `1 <= sz <= 30`
- `0 <= Node.val <= 100`
- `1 <= n <= sz`
### [Hints]
提示 1
Maintain two pointers and update one with a delay of n steps.
## 思路
1. 删除链表的倒数第 `N` 个结点，等同于删除链表的正数第 `node_count - N` 个结点。
2. 先求出`node_count`。
3. 在 `index == node_count - N` 时，进行删除节点操作：`node.next = node.next.next`。
4. 由于删除的节点可能是 `head`，所以使用虚拟节点 `dummy_node`，方便统一处理。
## 步骤
1. 求出`node_count`。
```ruby
node_count = 0
node = head
while node
node_count += 1
node = node.next
end
```
2. 在 `index == node_count - N`时，进行删除节点操作：`node.next = node.next.next`。
```ruby
index = 0
node = head
while node
if index == node_count - n
node.next = node.next.next
break
end
index += 1
node = node.next
end
```
3. 由于删除的节点可能是`head`，所以使用虚拟节点`dummy_node`，方便统一处理。
```ruby
dummy_head = ListNode.new # 1
dummy_head.next = head # 2
node = dummy_head # 3
# omitted code
return dummy_head.next
```
## 复杂度
- 时间复杂度: `O(N)`.
- 空间复杂度: `O(1)`.
## Java
```java
/**
* public class ListNode {
* int val;
* ListNode next;
* ListNode() {}
* ListNode(int val) { this.val = val; }
* ListNode(int val, ListNode next) { this.val = val; this.next = next; }
* }
*/
class Solution {
public ListNode removeNthFromEnd(ListNode head, int n) {
var nodeCount = 0;
var node = head;
while (node != null) {
nodeCount++;
node = node.next;
}
var index = 0;
var dummyHead = new ListNode(0, head);
node = dummyHead;
while (node != null) {
if (index == nodeCount - n) {
node.next = node.next.next;
break;
}
index++;
node = node.next;
}
return dummyHead.next;
}
}
```
## Python
```python
# Definition for singly-linked list.
# class ListNode:
# def __init__(self, val=0, next=None):
# self.val = val
# self.next = next
class Solution:
def removeNthFromEnd(self, head: Optional[ListNode], n: int) -> Optional[ListNode]:
node_count = 0
node = head
while node:
node_count += 1
node = node.next
index = 0
dummy_head = ListNode(next=head)
node = dummy_head
while node:
if index == node_count - n:
node.next = node.next.next
break
index += 1
node = node.next
return dummy_head.next
```
## C++
```cpp
/**
* Definition for singly-linked list.
* struct ListNode {
* int val;
* ListNode *next;
* ListNode() : val(0), next(nullptr) {}
* ListNode(int x) : val(x), next(nullptr) {}
* ListNode(int x, ListNode *next) : val(x), next(next) {}
* };
*/
class Solution {
public:
ListNode* removeNthFromEnd(ListNode* head, int n) {
auto node_count = 0;
auto node = head;
while (node != nullptr) {
node_count += 1;
node = node->next;
}
auto index = 0;
auto dummy_head = new ListNode(0, head);
node = dummy_head;
for (auto i = 0; i < node_count - n; i++) {
node = node->next;
}
auto target_node = node->next;
node->next = node->next->next;
delete target_node;
auto result = dummy_head->next;
delete dummy_head;
return result;
}
};
```
## JavaScript
```javascript
/**
* function ListNode(val, next) {
* this.val = (val===undefined ? 0 : val)
* this.next = (next===undefined ? null : next)
* }
*/
var removeNthFromEnd = function (head, n) {
let nodeCount = 0
let node = head
while (node != null) {
nodeCount++
node = node.next
}
let index = 0
let dummyHead = new ListNode(0, head)
node = dummyHead
while (node != null) {
if (index == nodeCount - n) {
node.next = node.next.next
break
}
index++
node = node.next
}
return dummyHead.next
};
```
## C#
```csharp
/**
* Definition for singly-linked list.
* public class ListNode {
* public int val;
* public ListNode next;
* public ListNode(int val=0, ListNode next=null) {
* this.val = val;
* this.next = next;
* }
* }
*/
public class Solution
{
public ListNode RemoveNthFromEnd(ListNode head, int n)
{
int nodeCount = 0;
var node = head;
while (node != null)
{
nodeCount++;
node = node.next;
}
int index = 0;
var dummyHead = new ListNode(0, head);
node = dummyHead;
while (node != null)
{
if (index == nodeCount - n)
{
node.next = node.next.next;
break;
}
index++;
node = node.next;
}
return dummyHead.next;
}
}
```
## Go
```go
/**
* Definition for singly-linked list.
* type ListNode struct {
* Val int
* Next *ListNode
* }
*/
func removeNthFromEnd(head *ListNode, n int) *ListNode {
nodeCount := 0
node := head
for node != nil {
nodeCount++
node = node.Next
}
index := 0
dummyHead := &ListNode{0, head}
node = dummyHead
for node != nil {
if index == nodeCount - n {
node.Next = node.Next.Next
break
}
index++
node = node.Next
}
return dummyHead.Next
}
```
## Ruby
```ruby
# class ListNode
# attr_accessor :val, :next
#
# def initialize(val = 0, _next = nil)
# @val = val
# @next = _next
# end
# end
def remove_nth_from_end(head, n)
node_count = 0
node = head
while node
node_count += 1
node = node.next
end
index = 0
dummy_head = ListNode.new(0, head)
node = dummy_head
while node
if index == node_count - n
node.next = node.next.next
break
end
index += 1
node = node.next
end
dummy_head.next
end
```
## Other languages
```java
// Welcome to create a PR to complete the code of this language, thanks!
```
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原文链接：[19. 删除链表的倒数第 N 个结点 - LeetCode Python/Java/C++/JS/C#/Go/Ruby 题解](https://leetcode.to/zh/leetcode/19-remove-nth-node-from-end-of-list).
GitHub 仓库: [leetcode-python-java](https://github.com/leetcode-python-java/leetcode-python-java).